All categories

2 years ago

Which of the following is NOT common of elite Shang burials?

Physics

1 answer:

Zigmanuir [339]2 years ago

6 0

Answer:

Large above ground mausoleums were not common in the elite Shang burials.

Explanation:

Large, above the ground mausoleums were not common so the answer is option B.

You might be interested in

Starting from rest, a truck travels in a straight line for 8.0 s with a uniform acceleration of +1.6 m/s2. The driver then appli

lana [24]

Because it should be the total number

4 0

4 years ago

What was the greatest discovery by galileo during his inclined-plane experiments?

Sergeeva-Olga [200]

Galileo discovered during his inclined-plane experiments that a ball rolling down an incline and onto a horizontal surface would roll indefinitely.

3 0

3 years ago

Order the speed of sound through these materials from the slowest to the fastest.

Sholpan [36]

Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Explanation:

Step 1:

Speed of sound in water varies from 1450 to 1498 meters per second

Speed of sound in Hot Molten lead is approximately 1210 meters per second

Speed of sound in warm air is approximately 338.89 meters per second

Speed of sound in cold air is approximately 293.33 meters per second

Step 2:

In warm air sound travels faster than that of sound travelling nature in cold air.

∴ Speed of sound in cold air < Speed of sound in Warm air < Speed of sound in hot molten lead < Speed of sound in water

Speed of sound in cold air the slowest while Speed of sound in water is the fastest mean.

8 0

3 years ago

vA 61.2-kg circus performer is fired from a cannon that is elevated at an angle of 57.8 ° above the horizontal. The cannon uses

dsp73

Answer:

The effective spring constant of the firing mechanism is 1808N/m.

Explanation:

First, we can use kinematics to obtain the initial velocity of the performer. Since we know the angle at which he was launched, the horizontal distance and the time in which it's traveled, we can calculate the speed by:

$v_0_x=\frac{x}{t}\\ \\v_0\cos\theta=\frac{x}{t}\\\\v_0=\frac{x}{t\cos\theta}$

(This is correct because the horizontal motion has acceleration zero). Then:

$v_0=\frac{20.8m}{(2.60s)\cos57.8\°}\\\\v_0=15.0m/s$

Now, we can use energy to obtain the spring constant of the firing mechanism. By the conservation of mechanical energy, considering the instant in which the elastic band is at its maximum stretch as t=0, and the instant in which the performer flies free of the bands as final time, we have:

$E_0=E_f\\\\U_e=K\\\\\frac{1}{2}kx^2=\frac{1}{2}mv^2\\\\\implies k=\frac{mv^2}{x^2}$

Then, plugging in the given values, we obtain:

$k=\frac{(61.2kg)(15.0m/s)^2}{(2.76m)^2}\\\\k=1808N/m$

Finally, the effective spring constant of the firing mechanism is 1808N/m.

3 0

3 years ago

A 7950-kg railroad car travels alone on a level frictionless track with a constant speed of 15.0 m/s . A 2950-kg load, initially

Alchen [17]

The new speed of car is 10.9 m/s

<h3 />

According to the principle of momentum conservation, momentum is only modified by the action of forces as they are outlined by Newton's equations of motion; momentum is never created nor destroyed inside a problem domain.

Mass of the railroad car, m₁ = 7950 kg

Mass of the load, m₂ = 2950 kg

It can be assumed as the speed of the car, u₁ = 15 m/s

Initially, it is at rest, u₂ = 0

Let v is the speed of the car. It can be calculated using the conservation of momentum as :

$m_1u_1 + m_2u_2 = (m_1 + m_2) v$